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Full text of "Treatise On Analysis Vol-Ii"

11    THE SPACES L1 AND L2       161

To show that Lg is complete, it is clearly enough to prove (i) and (ii). We
have already proved (i) in the case p = I (13.8.5). When p = 2, it follows from

00

the hypotheses and (13.11.3.1) that the function ]T \gn\ is finite almost every-

n= 1

00

where (13.6.4), and hence the series £ \gn(x)\ converges in R for almost all

n= 1

00                                                                                  00

x e X. Hence the same is true for £ gn(x\ and since \g(x)\ g £ \0n(x)\ almost

n=1                                                      «= 1

everywhere, we have N2(#) < +00 by (13.11.3.1). Moreover, g is measurable
(13.9.11), hence its class belongs to LR. Since

000 ~

^ I

for all n and almost all x e X, it follows fr6m (13.11.3.1) that

which is arbitrarily small for large n. Hence g = £ gn.

n = I

(ii)   To prove the first assertion, it is enough to show that, for a suitably

oo

chosen subsequence (/nfc), the series fni (x) +  £ (/»fc+i(*) -/»*(*)) converges

almost everywhere, and that if /(x) is its sum (defined almost everywhere)
then/ = lim /„. We define the sequence (nk) inductively to satisfy

for all k ^ 1 (and nv = 1); then it follows from (i) that the series

/.,<*)+ Z (/*„(«) -/„,(*))

fc=i
converges almost everywhere and that, if f(x) is its sum, then/=lim/Wk

fc->oo

in Lg. Since (/„) is a Cauchy sequence, it follows that /= lim /„ in Lg,

If"* 00

by (3.14.2). To prove the second assertion, the preceding argument maybe
applied by replacing the sequence (/„) by any subsequence (fni) which has a
limit almost everywhere.

(iii)   For each n let gn =    sup    \fn+h -fn+t\. Then gn is ;> 0 and meas-

urable (13.9.11), and belongs to J$?g~ (13.9.1 3) because \gn\ ^ 2/7. The sequence
(gn) is decreasing and by hypothesis tends to 0 almost everywhere, hence